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<meta property="og:description" content="最近推了一个圆锥曲线的结论, 因此记录一下. 先上结论: 若有圆锥曲线 $\Gamma\colon\frac{x^2}{A}+\frac{y^2}{B}&#x3D;1$ ,$A$ 与 $B$ 不同时取负数 (这样就同时代表了双曲线和椭圆) , 那么对于一个在该圆锥曲线上的点 $P_1(x_1,y_1)$ 来说, 存在一个点 $P_2(\frac{A-B}{A+B}x_1,\frac{B-A}{A+B}y_1">
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fa-fw"></i></a></div></div></nav><div id="post-info"><h1 class="post-title">圆锥曲线二级结论</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2021-03-21T03:32:14.000Z" title="发表于 2021-03-21 11:32:14">2021-03-21</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-08-14T04:39:34.408Z" title="更新于 2021-08-14 12:39:34">2021-08-14</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right post-meta-separator"></i><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%95%B0%E5%AD%A6/%E5%85%B6%E4%BB%96/">其他</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-pv-cv"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span><span class="post-meta-separator">|</span><span class="post-meta-commentcount"><i class="far fa-comments fa-fw post-meta-icon"></i><span class="post-meta-label">评论数:</span><a href="/math/other/conical_section_conclusion/#post-comment"><span id="twikoo-count"></span></a></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><p>最近推了一个圆锥曲线的结论, 因此记录一下.</p>
<p>先上结论:</p>
<p>若有圆锥曲线 $\Gamma\colon\frac{x^2}{A}+\frac{y^2}{B}=1$ ,$A$ 与 $B$ 不同时取负数 (这样就同时代表了双曲线和椭圆) , 那么对于一个在该圆锥曲线上的点 $P_1(x_1,y_1)$ 来说, 存在一个点 $P_2(\frac{A-B}{A+B}x_1,\frac{B-A}{A+B}y_1)$ . 若作直线 $l_1, l_2$ 过点 $P_0$ 且相互垂直, 那么这两条直线与该圆锥曲线异于 $P_0$ 的两个交点 $M,N$ 的连线恒过点 $P_1$ .</p>
<iframe 
frameborder="no"
title="垂点椭圆双曲线" src="https://www.geogebra.org/material/iframe/id/zkwudkkg/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/true/rc/false/ld/false/sdz/true/ctl/false/" 
width=100% 
height="480px">
</iframe>



<p>对于形如 $y^2=2px$ 的抛物线也有类似结论, 不过 $P_2$ 为 $(2p+x_1,-y_1)$.</p>
<iframe 
frameborder="no"
title="垂点抛物线" src="https://www.geogebra.org/material/iframe/id/myysrpxf/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/true/rc/false/ld/false/sdz/true/ctl/false/" 
width=100% 
height="480px">
</iframe>


<h1 id="如何得到"><a href="#如何得到" class="headerlink" title="如何得到"></a>如何得到</h1>

我们以椭圆为例.

设椭圆 $\Gamma\colon\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , 点 $P_2 (x_2, y_2)$ , 直线 $l_{MN}\colon y-y_2=k(x-x_2)$  联立可得
$$
(a^2k^2+b^2)x^2+2a^2k(y_2-kx_2)x+a^2((kx_2-y_2)^2-b^2)=0
$$
设直线 $l_{MN}$ 与曲线交点为 $M(x_m,y_m),N(x_n,y_n)$ 根据韦达定理可得
$$
\begin{aligned}
x_mx_n=\frac{a^2((kx_2-y_2)^2-b^2)}{a^2k^2+b^2},x_m+x_n=\frac{2a^2k(kx_2-y_2)}{a^2k^2+b^2}\\
y_my_n=\frac{b^2((kx_2-y_2)^2-a^2k^2)}{a^2k^2+b^2},y_m+y_n=\frac{2b^2(y_2-kx_2)}{a^2k^2+b^2}
\end{aligned}
$$
同时设 $P_1(x_1,y_1)$ ,有
$$
\begin{aligned}
(x_m-x_1,y_m,y_1)\cdot(x_n-x_2,y_n-y_2)&=0\\
x_nx_m-x_1(x_m+x_n)+{x_2}^2+y_ny_m-y_1(y_m+y_n)+{y_2}^2&=0
\end{aligned}
$$
代入得
$$
\begin{aligned}
a^2((kx_2-y_2)^2-b^2)-2x_1a^2k(kx_2-y_2)+b^2((kx_2-y_2)^2-a^2k^2)-2y_1b^2(y_2-kx_2)\\=-({x_1}^2+{y_1}^2)(a^2k^2+b^2)
\end{aligned}
$$
即上式对于任意 $k$ 成立. 凭借这个复杂的等式我们已可以 (理论上) 求出 $P_1$ 与 $P_2$ 的关系, 但是这样太困难了, 我们稍稍取下巧, 只需取两个 $k$ 值就行, 很容易想到 $k=0$ 与 $k\to\infty$ (即令 $1/ \to 0$),

令 $k=0$ 得
$$
a^2({y_2}^2-b^2)+b^2{y_2}^2-2b^2y_1y_2=-b^2({x_1}^2+{y_1}^2)
$$
令 $k\to\infty$ 得
$$
b^2({x_2}^2-a^2)+a^2{x_2}^2-2a^2x_1x_2=-a^2({x_1}^2+{y_1}^2)
$$
这就好多了. 但是我们发现, 已知 $x_2,y_2$ 去求 $x_1,y_1$ 有困难, 但是我们稍稍整理一下就会发现
$$
(a^2+b^2)\cdot {y_2}^2 -2b^2y_1\cdot y_2 + b^2({x_1}^2+{y_1}^2)-a^2b^2=0\qquad(k=0)
$$
而这恰好是一个以 $x_1,y_1$ 作为参数的关于 $y_2$ 的二元一次方程, 那么就让我们把 $x_1,y_1$ 看成已知, 而 $x_2,y_2$ 看成未知的吧.

用求根公式可得
$$
y_2=\frac{2b^2y_1\pm\sqrt{4b^2{y_1}^2-4(a^2+b^2)(b^2({x_1}^2+{y_1}^2)-a^2b^2)}}{2(a^2+b^2)}
$$
$x_2$ 的求法也是同理 (利用 $k\to \infty$ 的情况) , 这样我们就得出了结果.

等会... 这个公式和上面那个完全不一样啊! 而且还带 $\pm$ , 显然有一个解得舍去的啊! 事实上, 最上面那个公式只是这个公式的简化, 运用了到目前还没用过的最重要的等式, 即
$$
\begin{aligned}
\frac{{x_1}^2}{a^2}+\frac{{y_1}^2}{b^2}&=1\\
a^2-{x_1}^2&=\frac{a^2}{b^2}{y_1}^2
\end{aligned}
$$
因此有
$$
\begin{aligned}
y_2&=\frac{2b^2y_1\pm\sqrt{4b^2{y_1}^2-4(a^2+b^2)(b^2({x_1}^2+{y_1}^2)-a^2b^2)}}{2(a^2+b^2)}\\
&=\frac{b^2y_1\pm\sqrt{b^2(a^2-{x_1}^2)(a^2+b^2)-a^2b^2{y_1}^2}}{a^2+b^2}\\
&=\frac{b^2y_1\pm\sqrt{a^2{y_1}^2(a^2+b^2)-a^2b^2{y_1}^2}}{a^2+b^2}\\
&=\frac{b^2y_1\pm a^2y_1}{a^2+b^2}
\end{aligned}
$$
取 $+$ 的不符合实际情况 $y_1=y_2$ , 于是就有
$$
y_1=\frac{b^2-a^2}{a^2+b^2}y_2
$$
同理得
$$
x_1=\frac{a^2-b^2}{a^2+b^2}x_2
$$
双曲线和椭圆公式类似, 可以用同样方法求得, 并且归入最上面说到的一整个结论.

但是抛物线和它们可不太一样, 难道要重新求一遍? 并不, 注意到抛物线是椭圆的一种极限. 设抛物线 $E\colon y^2=2px$ , 有
$$
E\colon\lim_{n\to\infty} \frac{(x-np)^2}{n^2p^2}+\frac{y^2}{np^2}=1
$$
利用上面求出的公式, 有
$$
\begin{aligned}
x_2&=\lim_{n\to\infty}np+(x-np)\frac{n^2p^2-np^2}{n^2p^2+np^2}\\
&=\lim_{n\to\infty}np+(x-np)\frac{n-1}{n+1}\\
&=\lim_{n\to\infty}\frac{n-1}{n+1}x+\frac{2np}{n+1}\\
&=2p+x_1
\end{aligned}
$$

$y_2=-y_1$ 也是类似求法.

由上得到了全部结果.



<h1 id="感想"><a href="#感想" class="headerlink" title="感想"></a>感想</h1><p>解析几何就一句话: 干就完事了!</p>
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